# Chapter 5: Factorization of Algebraic Expressions Exercise – 5.1

Find the value

x3 + x − 3x2 − 3

### Solution:

Taking x common in x3 + x

= x(x2 + 1) − 3x2 − 3

Taking - 3 common in − 3x2 − 3

= x(x2 + 1) − 3(x2 + 1)

Now, we take (x2 + 1) common

= (x2 + 1) (x - 3)

∴ x3 + x − 3y2 − 3 = (x2 + 1)(x - 3)

### Question: 2

Find the value

a(a + b)3 − 3a2b(a + b)

### Solution:

Taking (a + b) common in the two terms

= (a + b){a(a + b)² - 3a²b}

Now, using (a + b)2= a2 + b2 + 2ab

= (a + b){a(a2 + b2 + 2ab) − 3a2b}

= (a + b){a3 + ab2 + 2a2b − 3a2b}

= (a + b){a3 + ab2 − a2b}

= (a + b)p{a2 + b2 − ab}

= p(a + b)(a2 + b2 − ab)

∴ a(a + b)3 − 3a2b(a + b)

= a(a + b)(a2 + b2 − ab)

### Question: 3

Find the value

x(x3 − y3) + 3xy(x − y)

### Solution:

Elaborating x3 − y3 using the identity

x3 − y3 = (x − y)(x2 + xy + y2)

= x(x − y)(x2 + xy + y2) + 3xy(x − y)

Taking common x(x - y) in both the terms

= x(x − y)(x2 + xy + y2 + 3y)

∴ x(x3 − y3) + 3xy(x − y)

= x(x − y)(x2 + xy + y2 + 3y)

### Question: 4

Find the value

a2x2 + (ax2 + 1)x + a

### Solution:

We multiply x(ax2 + 1) = ax3 + x

= a2x2 + ax3 + x + a

Taking common ax2 in (a2x2 + ax3) and 1 in (x + a)

= ax2(a + x) + 1(x + a)

=ax2(a + x) + 1(a + x)

Taking (a + x) common in both the terms

= (a + x)(ax2 + 1)

∴ a2x2 + (ax2 + 1)x + a = (a + x)(ax2 + 1)

Find the value

x2 + y − xy − x

### Solution:

On rearranging

x2 − xy − x + y

Taking x common in the (x2 − xy) and -1 in(-x + y)

= x(x - y) - 1 (x - y)

Taking (x - y) common in the terms

= (x - y)(x - 1)

∴ x2 + y − xy − x = (x - y)(x - 1)

### Question: 6

Find the value

x3 − 2x2b + 3xy2 − 6y3

### Solution:

Taking x2 common in (x3 − 2x2y) and +3y2 common in (3xy2 − 6y3)

= x2(x − 2y) + 3y2(x − 2y)

Taking (x - 2y) common in the terms

= (x − 2y)(x2 + 3y2)

∴ x3 − 2x2y + 3xy2 − 6y3 = (x − 2y)(x2 + 3y2)

### Question: 7

Find the value

6ab − b2 + 12ac − 2bc

### Solution:

Taking b common in (6ab − b2) and 2c in (12ac - 2bc)

= b(6a - b) + 2c (6a - b)

Taking (6a - b) common in the terms

= (6a - b)(b + 2c)

∴ 6ab − b2 + 12ac − 2bc = (6a - b)(b + 2c)

### Question: 8

Find the value

[x2 + 1/x2] − 4[x + 1/x] + 6

### Solution:

= x2 + 1/x2 − 4x − 4/x + 4 + 2

= x2 + 1/x2 + 4 + 2 − 4x − 4x

= (x2) + (1/x)2 + (−2)2 + 2 × x × 1/x + 2 × 1/x × (−2) + 2(−2)x

Using identity

x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2

We get,

= [x + 1/x + (−2)]2

= [x + 1/x − 2]2

= [x + 1/x − 2][x + 1/x − 2]

∴ [x2 + 1/x2] − 4[x + 1/x] + 6 = [x + 1/x − 2][x + 1/x − 2]

### Question: 9

Find the value

x(x - 2)(x - 4) + 4x - 8

### Solution:

= x(x  - 2)(x - 4) + 4(x - 2)

Taking (x - 2) common in both the terms

=(x - 2){x(x - 4) + 4}

=(x - 2){x2 − 4x + 4}

Now splitting the middle term of x2 − 4x + 4

= (x - 2){x2 − 2x − 2x + 4}

= (x - 2){x( x - 2) -2(x - 2)}

= (x - 2){(x - 2)(x - 2)}

= (x - 2)(x - 2)(x - 2)

= (x − 2)3

∴ x(x - 2)(x - 4) + 4x - 8 = (x − 2)3

### Question: 10

Find the value

(x + 2)(x2 + 25) − 10x2 − 20x

### Solution:

(x + 2)(x2 + 25) - 10x (x + 2)

Taking (x + 2) common in both the terms

= (x + 2)(x2 + 25 − 10x)

= (x + 2)(x2 − 10x + 25)

Splitting the middle term of (x2 − 10x + 25)

= (x + 2)(x2 − 5x − 5x + 25)

= (x + 2){x(x - 5)-5 (x - 5)}

= (x + 2)(x - 5)(x - 5)

∴ (x + 2)(x2 + 25) − 10x2 - 20x = (x + 2)(x - 5)(x - 5)

Find the value

2a2+2√6ab +3b2

### Solution:

Using the identity (p + q)2 = p2 + q2 + 2pq

### Question: 12

Find the value

(a − b + c)2 + (b − c + a)2 + 2(a − b + c) × (b − c + a)

### Solution:

Let (a - b + c) = x and (b - c + a) = y

= x2 + y2 + 2xy

Using the identity (a + b)2 = a2 + b2 + 2ab

= (x + y)2

Now, substituting  x and y

(a - b + c + b − c + a)2

Cancelling -b, +b  & + c, -c

= (2a)2

= 4a2

∴ (a − b + c)2 + (b − c + a)2 + 2(a − b + c) × (b − c + a) = 4a2

### Question: 13

Find the value

a2 + b2 + 2(ab + bc + ca)

### Solution:

= a2 + b2 + 2ab + 2bc + 2ca

Using the  identity (p + q)2 = p2 + q2 + 2pq

We get,

= (a + b)2 + 2bc + 2ca

= (a + b)2 + 2c(b + a)

Or (a + b)2 + 2c(a + b)

Taking (a + b) common

= (a + b)(a + b + 2c)

∴ a2 + b2 + 2(ab + bc + ca) = (a + b)(a + b + 2c)

### Question: 14

Find the value

4(x − y)2 − 12(x − y)(x + y) + 9(x + y)2

### Solution:

Let(x - y) = x,(x + y) = y

= 4x2 − 12xy + 9y2

Splitting the middle term - 12 = – 6 - 6  also 4 × 9 = −6 × − 6

= 4x2 − 6xy − 6xy + 9y2

= 2x(2x - 3y) -3y(2x - 3y)

= (2x - 3y)(2x - 3y)

= (2x − 3y)2

Substituting x = x - y & y = x + y

= [2(x − y) − 3(x + y)]2 = [2x - 2y - 3x - 3y]2

= (2x - 3x - 2y - 3y)²

= [−x − 5y]2

= [(−1)(x + 5y)]2

= (x + 5y)2    [? (-1)2 = 1]

∴ 4(x − y)2 − 12(x − y)(x + y) + 9(x + y)2 = (x + 5y)2

### Question: 15

Find the value

a2 − b2 + 2bc − c2

### Solution:

a2 − (b2 − 2bc + c2)

Using the identity (a − b)2 = a2 + b2 − 2ab

= a2 − (b − c)2

Using the identity a2 − b2 = (a + b)(a − b)

= (a + b - c)(a - (b - c))

= (a + b - c)(a - b + c)

∴ a2 − b2 + 2bc − c2 = (a + b - c)(a - b + c)

### Question: 16

Find the value

a2 + 2ab + b2 − c2

### Solution:

Using the identity (p + q)2 = p2 + q2 + 2pq

= (a + b)2 − c2

Using the identity p2 − q2 = (p + q)(p − q)

= (a + b + c)(a + b - c)

∴ a2 + 2ab + b2 − c2 = (a + b + c)(a + b - c)

### Question: 17

Find the value

a2 + 2ab + b2 − c2

### Solution:

Using the identity (p + q)2 = p2 + q2 + 2pq

= (a + b)2 − c2

Using the identity p2 − q2 = (p + q)(p − q)

= (a + b + c)(a + b - c)

∴  a2 + 2ab + b2 − c2 = (a + b + c)(a + b - c)

Find the value

xy9 − yx9

### Solution:

= xy(y8 − x8)

= xy((y4)2 − (x4)2)

Using the identity p2 − q2 = (p + q)(p - q)

= xy(y4 + x4)(y4 − x4)

= xy(y4 + x4)((y2)2 − (x2)2)

Using the identity p2 − q2 = (p + q)(p - q)

= xy(y4 + x4)(y2 + x2)(y2 − x2)

= xy(y4 + x4)(y2 + x2)(y + x)(y − x)

= xy(x4 + y4)(x2 + y2)(x + y)(−1)(x − y)

∴ (y − x) = −1(x − y)

= −xy(x4 + y4)(x2 + y2)(x + y)(x − y)

∴ xy9 − yx9 = −xy(x4 + y4)(x2 + y2)(x + y)(x − y)

Find the value

x4 + x2y2 + y4

### Solution:

Adding x2y2 and subtracting x2y2 to the given equation

= x4 + x2y2 + y4 + x2y2 − x2y2

= x4 + 2x2y2 + y4 − x2y2

= (x2)2 + 2 × x2 × y2 + (y2)2 − (xy)2

Using the identity (p + q)2 = p2 + q2 + 2pq

= (x2 + y2)2 − (xy)2

Using the identity  p2 − q2 = (p + q)(p - q)

= (x2 + y2 + xy)(x2 + y2 − xy)

∴  x4 + x2y2 + y4 = (x2 + y2 + xy)(x2 + y2 − xy)

### Question: 20

Find the value

x2 − y2 − 4xz + 4z2

### Solution:

On rearranging the terms

= x2 − 4xz + 4z2 − y2

= (x)2 − 2 × x × 2z + (2z)2 − y2

Using the identity x2 − 2xy + y2 = (x − y)2

= (x − 2z)2 − y2

Using the identity p2 − q2 = (p + q)(p - q)

= (x − 2z + y)(x − 2z − y)

∴ x2 − y2 − 4xz + 4z2 = (x − 2z + y)(x − 2z − y)

Find the value

### Solution:

Splitting the middle term,

Find the value

### Solution:

Splitting the middle term,

Find the value

### Solution:

Splitting the middle term,

Find the value

### Solution:

Splitting the middle term,

Find the value

### Solution:

Splitting the middle term,

Find the value

### Solution:

Splitting the middle term,

= 2x2 − x2 − x3 + 1/12

[∴ −5/6 = −1/2 − 1/3 also −1/2 × −1/3 = 2 × 1/12]

= x(2x − 1/2) − 1/6(2x − 1/2)

= (2x − 1/2)(x − 1/6)

∴ 2x2 − 56x + 1/12 = (2x − 1/2)(x − 1/6)

Find the value

### Solution:

Splitting the middle term,

= x2 + x/7 + x/5 + 1/35

= x(x + 1/7) + 1/5(x + 1/7)

= (x + 1/7)(x + 1/5)

Find the value

21x2 − 2x + 1/21

### Solution:

Using the identity (x - y)2 = x2 + y2 - 2xy

Find the value

### Solution:

Splitting the middle term,

Find the value

### Solution:

Splitting the middle term,

### Question: 31

Find the value

9(2a − b)2 − 4(2a − b) − 13

### Solution:

Let 2a - b = x

= 9x2 − 4x − 13

Splitting the middle term,

= 9x2 − 13x + 9x − 13

= x(9x − 13) + 1(9x − 13)

= (9x − 13)(x + 1)

Substituting x = 2a - b

= [9(2a − b) − 13](2a − b + 1)

= (18a − 9b − 13)(2a − b + 1)

∴ 9(2a − b)2 − 4(2a − b) − 13 = (18a − 9b − 13)(2a − b + 1)

### Question: 32

Find the value

7(x − 2y)2 − 25(x − 2y) + 12

### Solution:

Let x - 2y = P

= 7P2 − 25P + 12

Splitting the middle term,

= 7P2 − 21P − 4P + 12

= 7P(P − 3) − 4(P − 3)

= (P − 3)(7P − 4)

Substituting  P = x - 2y

= (x − 2y − 3)(7(x − 2y) − 4)

= (x − 2y − 3)(7x − 14y − 4)

∴ 7(x − 2y)2 − 25(x − 2y) + 12 = (x − 2y − 3)(7x − 14y − 4)

### Question: 33

Find the value

2(x + y)2 − 9(x + y) − 5

### Solution:

Let x + y = z

= 2z2 − 9z − 5

Splitting the middle term,

= 2z2 − 10z + z − 5

= 2z(z − 5) + 1(z − 5)

= (z − 5)(2z + 1)

Substituting z = x + y

= (x + y − 5)(2(x + y) + 1)

= (x + y − 5)(2x + 2y + 1)

∴ 2(x + y)2 − 9(x + y) − 5 = (x + y − 5)(2x + 2y + 1)

### Question: 34

Give the possible expression for the length & breadth of the rectangle having 35y2 − 13y − 12 as its area.

### Solution:

Area is given as 35y2 − 13y − 12

Splitting the middle term,

Area =  35y2 + 218y − 15y − 12

= 7y(5y + 4) − 3(5y + 4)

= (5y + 4)(7y − 3)

We also know that area of rectangle = length × breadth

∴ Possible length = (5y + 4) and breadth = (7y − 3)

Or possible length = (7y − 3) and breadth = (5y + 4)

### Question: 35

What are the possible expression for the cuboid having volume 3x2 − 12x.

### Solution:

Volume = 3x2 − 12x

= 3x(x − 4)

= 3×x(x − 4)

Also volume = Length × Breadth × Height

∴ Possible expression for dimensions of cuboid are = 3, x, (x − 4)

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